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2 years ago

The process of how rocks can tell us how old the earth is

1 answer:

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The Earth is a constantly changing planet. Its crust is continually being created, modified, and destroyed. As a result, rocks that record its earliest history have not been found and probably no longer exist. Nevertheless, there is substantial evidence that the Earth and the other bodies of the Solar System are 4.5-4.6 billion years old, and that the Milky Way Galaxy and the Universe are older still. The principal evidence for the antiquity of Earth and its cosmic surroundings is:The oldest rocks on Earth, found in western Greenland, have been dated by four independent radiometric dating methods at 3.7-3.8 billion years. Rocks 3.4-3.6 billion years in age have been found in southern Africa, western Australia, and the Great Lakes region of North America. These oldest rocks are metamorphic rocks but they originated as lava flows and sedimentary rocks. The debris from which the sedimentary rocks formed must have come from even older crustal rocks. The oldest dated minerals (4.0-4.2 billion years) are tiny zircon crystals found in sedimentary rocks in western Australia.
The oldest Moon rocks are from the lunar highlands and were formed when the early lunar crust was partially or entirely molten. These rocks, of which only a few were returned by the Apollo missions, have been dated by two methods at between 4.4-4.5 billion years in age.
The majority of the 70 well-dated meteorites have ages of 4.4-4.6 billion years. These meteorites, which are fragments of asteroids and represent some of the most primitive material in the solar system, have been dated by 5 independent radiometric dating methods.
The "best" age for the Earth is based on the time required for the lead isotopes in four very old lead ores (galena) to have evolved from the composition of lead at the time the Solar System formed, as recorded in the Canyon Diablo iron meteorite. This "model lead age" is 4.54 billion years.
The evidence for the antiquity of the Earth and Solar System is consistent with evidence for an even greater age for the Universe and Milky Way Galaxy. a) The age of the Universe can be estimated from the velocity and distance of galaxies as the universe expands. The estimates range from 7 to 20 billion years, depending on whether the expansion is constant or is slowing due to gravitational attraction. b) The age of the Galaxy is estimated to be 14-18 billion years from the rate of evolution of stars in globular clusters, which are thought to be the oldest stars in the Galaxy. The age of the elements in the Galaxy, based on the production ratios of osmium isotopes in supernovae and the change in that ratio over time due to radioactive decay, is 8.6-15.7 billion years. Theoretical considerations indicate that the Galaxy formed within a billion years of the beginning of the Universe. c) Combining the data from a) and b), the "best, i.e., most consistent, age of the universe is estimated to be around 14 billion years. For more current information on the age of the universe.

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The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the

Oksana_A [137]

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t

⇒ α = (0 rev/min - 113 rev/min) / (60 min)

⇒ α = - 1.883 rev/min²

6 0

3 years ago

A box is being pulled to the right. What is the magnitude of the Kinect frictional force?

Anna35 [415]

The answer to this question is A - 25 N

3 0

2 years ago

Read 2 more answers

A 1500 kg car, initially traveling at 22.0 m/s, hits its brakes and skids to a stop. Determine the work done by friction.

Sedbober [7]

Answer:

The work done by the car is 363 kJ

Explanation:

Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).

Mathematically work done can be expressed as,

E = W = 1/2mv²

W = 1/2mv²................................ Equation 1

Where E = Energy, W = work done, m = mass of the car, v = velocity of the car

Given: m=1500 kg, v=22 m/s

Substituting these values into equation 1

W = 1/2(1500)(22)²

W = 750 × 484

W = 363000 J

W = 363 kJ

Thus the work done by the car is 363 kJ

8 0

3 years ago

mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi

valina [46]

Answer:

The answer is

" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

$W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\$

The source of the hooks law is stable,

$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

Assume that solution equation,

$x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\$

These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

Going to compare cos3 t and sin 3 t coefficients from both sides,

The cost3 t is 3A + 3B= 20 coefficients

The sin 3 t is 3B -3A = 0 coefficient

The two equations solved:

$3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\$

Replace the very first equation with the meaning,

$3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\$

equation is

$x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)$

The ultimate plan for both the equation is therefore

$x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)$

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

$x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\$

$x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\$

$c_2=\frac{-64\sqrt{47}}{141}$

$x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$

5 0

3 years ago

If you double the wavelength, the electromagnetic radiation energy will double. The energy of the electromagnetic radiation will

Sedbober [7]

Answer:

It's energy will double.

Explanation:

This is because energy, E, is related to frequency, f, by:

E = hf

Where h = Planck's constant

So, double frequency will be 2f

=> E(2f) = 2hf = 2E.

Hence, energy is doubled.

7 0

3 years ago